Complex Numbers Q (1 Viewer)

nrlwinner · Jan 25, 2010

Show that the equation x^2 +kx +1 = 0 has complex roots with unit modulus provided -2<k<2

Trebla · Jan 25, 2010

nrlwinner said:
Show that the equation x^2 +kx +1 = 0 has complex roots with unitmodulus provided -2<k<2

$$Since polynomial is assumed to be real, the roots must occur in conjugate pairs say $z$ and $ \bar{z}\\\\$Product of roots $\quad z\bar{z}=\left |z \right |^2=1\\\\\Rightarrow \left | z \right | = \left | \bar{z} \right | =1\\\\$Also, $ \Delta =k^2-4\\\\\Delta < 0 $ for non-real roots if $ -2 < k < 2$

nrlwinner · Jan 25, 2010

I don't get the discriminate part.

What relevance does it have?

Lukybear · Jan 25, 2010

Roots must not be real, for complex conjugates to occur I think. Thus unit modulous cannot occur, if roots were real.

Did you post the full quesiton and can you tell me where is it from?

nrlwinner · Jan 25, 2010

Lukybear said:
Roots must not be real, for complex conjugates to occur I think. Thus unit modulous cannot occur, if roots were real.

Did you post the full quesiton and can you tell me where is it from?

Yeah it's the full queston and you're right. I got the question from my tutor.

I've got another one here. I've partially solved it.

$\frac{1}{1+z}=\frac{1}{2}(1-itan\frac{\Theta }{2})$ I'm trying to prove that. I've got up to

$\frac{1}{2}(\frac{cos\frac{\Theta }{2}(cos\frac{\Theta }{2}-isin\frac{\Theta }{2})}{1-sin^2\frac{\Theta }{2}})$

Lukybear · Jan 25, 2010

way of my league thats for sure...

but in ur question what does "provided -2 " mean?

nrlwinner · Jan 25, 2010

Lukybear said:
way of my league thats for sure...

but in ur question what does "provided -2 " mean?

it was meant to be -2<k<2

Lukybear · Jan 25, 2010

do you mean the domain? or the root is -2

untouchablecuz · Jan 26, 2010

$\\$Let $z=\cos \theta +i\sin \theta $ and $t=\tan \frac{\theta}{2}:\\\frac{1}{1+z}\\=\frac{1}{1+\cos \theta+i\sin \theta}\\=\frac{1}{1+\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}}\\=\frac{1}{2} \times\frac{1+t^2}{1+it}\\=\frac{1}{2} \times\frac{(1+it)(1-it)}{1+it}\\=\frac{1}{2} \times(1-it)\\=\frac{1}{2} \times(1-i\tan\frac{\theta}{2})$

you can also do it using the identities,

$\cos \theta = 2\cos^2 \frac{\theta}{2} -1=1-2\sin^2 \frac{\theta}{2}$ which is more tedious

nrlwinner · Jan 26, 2010

untouchablecuz said:
$\\$Let $z=\cos \theta +i\sin \theta $ and $t=\tan \frac{\theta}{2}:\\\frac{1}{1+z}\\=\frac{1}{1+\cos \theta+i\sin \theta}\\=\frac{1}{1+\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}}\\=\frac{1}{2} \times\frac{1+t^2}{1+it}\\=\frac{1}{2} \times\frac{(1+it)(1-it)}{1+it}\\=\frac{1}{2} \times(1-it)\\=\frac{1}{2} \times(1-i\tan\frac{\theta}{2})$

you can also do it using the identities,

$\cos \theta = 2\cos^2 \frac{\theta}{2} -1=1-2\sin^2 \frac{\theta}{2}$ which is more tedious

Yeah, I was using identities and I always seem to be stuck. Can you try for me because it's the way I'v learnt to do it.

gurmies · Jan 26, 2010

nrlwinner · Jan 26, 2010

gurmies said:
$\frac{1}{2}\left [ \frac{\cos\frac{\theta}{2}\left ( \cos\frac{\theta}{2}-i\sin\frac{\theta}{2} \right )}{1-sin^{2}\frac{\theta}{2}} \right ]=\frac{1}{2}\left [ \frac{\cos\frac{\theta}{2}\left ( \cos\frac{\theta}{2}-i\sin\frac{\theta}{2} \right )}{\cos^{2}\frac{\theta}{2}} \right ] \\\\ = \frac{1}{2}\left ( 1-i\tan\frac{\theta}{2} \right )$

My god thanks. Cannot believe I missed that.

gurmies · Jan 26, 2010

nrlwinner said:
My god thanks. Cannot believe I missed that.

You were very close though, keep at it

untouchablecuz · Jan 26, 2010

nrlwinner · Jan 26, 2010

No idea how to do this one...

$If\ z=x+iy\ and\ z^2=a+ib$

Prove

$2x^2=\sqrt{(a^2+b^2)}+a$

Trebla · Jan 26, 2010

untouchablecuz · Jan 26, 2010

$\\z^2=a+ib\\|z^2|=|a+ib|\\|z|^2=|a+ib|\\(\sqrt{x^2+y^2})^2=\sqrt{a^2+b^2}\\x^2+y^2=\sqrt{a^2+b^2} \ (*)\\$Now, $\\(x+iy)^2=a+ib\\(x^2-y^2)+2ixy=a+ib\\$Equating $\Re (z),\\x^2-y^2=a\Rightarrow y^2=x^2-a \ (**)\\$Sub $(**) $ in $(*),\\2x^2=a+\sqrt{a^2+b^2}$

edit: beaten by the Treb Man

Complex Numbers Q (1 Viewer)

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